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LESSON 13.3<br />

1. Case 1: P is collinear with A and B. Use the Line<br />

Intersection Postulate to show that P and E are the<br />

same point and use the definitions of bisector and<br />

midpoint to get AP BP.<br />

Case 2: P is not collinear with A<br />

and B. Use the SAS Congruence Postulate to get<br />

AEP BEP. Then use CPCTC to get AP BP.<br />

A<br />

2. Case 1: P is collinear with A and B. Use the<br />

definitions of congruence and midpoint to show<br />

that P is the midpoint of AB.Then use the<br />

definition of perpendicular bisector.<br />

A<br />

C<br />

P<br />

E<br />

D<br />

P<br />

E<br />

Case 2: P is not collinear with A and B.Draw<br />

midpoint E and PE. Use the SSS Congruence<br />

Postulate to get AEP BEP. Then use CPCTC<br />

and the Congruent and Supplementary Theorem<br />

to prove that AEP and BEP are both right<br />

angles. Therefore PE is the perpendicular bisector<br />

of AB by the definitions of midpoint and<br />

perpendicular.<br />

3. Use the reflexive property and<br />

C<br />

the SSS Congruence Postulate to get<br />

ABC BAC. Therefore,<br />

A B by CPCTC.<br />

4. Use the reflexive property and<br />

the ASA Congruence Postulate to<br />

get ABC BAC. Then use<br />

CPCTC and the definition of<br />

isosceles triangle.<br />

5.<br />

B<br />

C A<br />

P<br />

B<br />

B<br />

A B<br />

A B<br />

Draw line BA.<br />

Use the Isosceles Triangle Theorem to get<br />

PAB PBA. Use the Angle Addition<br />

Postulate and the subtraction property to get<br />

BAC ABC. Then use the Converse of the<br />

C<br />

Isosceles Triangle Theorem (CB CA), the reflexive<br />

property, and the SSS Congruence Postulate to get<br />

ACP BCP. Therefore,ACP BCP by<br />

CPCTC.<br />

6.<br />

C m<br />

A<br />

Use the Line Intersection Postulate and the<br />

Perpendicular Bisector Theorem to get AP BP<br />

and BP CP. Then use the transitive property and<br />

the Converse of the Perpendicular Bisector<br />

Theorem to prove that point P is on line n.<br />

7.<br />

C<br />

Use the Line Intersection Postulate and the Angle<br />

Bisector Theorem to prove that Q is equally distant<br />

from AB and AC and from AB and BC. Then use the<br />

transitive property and the Converse of the Angle<br />

Bisector Theorem to prove that point Q is on line n.<br />

8.<br />

B<br />

A<br />

Use the Linear Pair Postulate and the definition of<br />

supplementary angles to get m3 m4 180°.<br />

Then use the Triangle Sum Theorem and the<br />

transitive property to get m1 m2 <br />

m3 m3 m4. Therefore, m1 <br />

m2 m4 by the subtraction property.<br />

9. D<br />

A<br />

n<br />

m<br />

A B<br />

1<br />

3 4<br />

2<br />

1<br />

2<br />

B<br />

Q<br />

n<br />

P<br />

<br />

3 4<br />

C<br />

C<br />

<br />

Use the Triangle Sum Theorem and the addition<br />

property to get mA m1 m3 mC <br />

m4 m2 360°. Then use the Angle Addition<br />

Postulate and the substitution property to get<br />

mA mABC mC mCDA 360°.<br />

10. Use the definitions of median<br />

and midpoint to get BM 1 BC 2 and<br />

AN 1<br />

C<br />

N M<br />

AC. 2 Then use the multiplication<br />

property and the substitution<br />

A B<br />

property to get AN BM . By the<br />

reflexive property, the Isosceles Triangle Theorem,<br />

and the SAS Congruence Postulate, ABN <br />

BAM. Therefore, BN AM by CPCTC.<br />

B<br />

ANSWERS TO EXERCISES 147<br />

<strong>Answers</strong> to Exercises

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